You have found the following ages (in years) of 6 porcupines. Those porcupines were randomly selected from the 45 porcupines at your local zoo: $ 3,\enspace 1,\enspace 24,\enspace 9,\enspace 23,\enspace 6$ Based on your sample, what is the average age of the porcupines? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 45 porcupines, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{3 + 1 + 24 + 9 + 23 + 6}{{6}} = {11\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {64} + {100} + {169} + {4} + {144} + {25}} {{6 - 1}} $ {s^2} = \dfrac{{506}}{{5}} = {101.2\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{101.2\text{ years}^2}} = {10.1\text{ years}} $ We can estimate that the average porcupine at the zoo is 11 years old. There is also a standard deviation of 10.1 years.